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The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. What is the number of page faults?
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370
- a)13
- b)8
- c)7
- d)10
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The address sequence generated by tracing a particular program executi...
When it tries to access 0100, it results in a page fault as the memory is empty right now. So, it loads the second page (which has the addresses 100 - 199). Trying to access 200 will result in a page fault, as it is not in memory right now. So the third page with the addresses from 200 to 299 will replace the second page in memory. Trying to access 430 will result in another page fault. Proceeding this way, we find trying to access the addresses 0510,0120,0220 and 0320 will all result in page faults. So, altogether 7 page faults.
Most Upvoted Answer
The address sequence generated by tracing a particular program executi...
Explanation:
Given:
Page size = 100 records
Free main memory frame = 1
Address Sequence:
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370
Calculation:
Initially, the main memory is empty. When the program starts executing, it requests the first page, which causes a page fault because the page is not in the main memory. The operating system then loads the page into the main memory.
When the program requests the second page (0200), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the first page.
Similarly, when the program requests the third page (0430), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the second page.
When the program requests the fourth page (0499), it also causes a page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the third page.
When the program requests the fifth page (0510), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the fourth page.
When the program requests the sixth page (0530), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the fifth page.
When the program requests the seventh page (0560), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the sixth page.
When the program requests the eighth page (0120), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the seventh page.
When the program requests the ninth page (0220), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the eighth page.
When the program requests the tenth page (0240), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the ninth page.
When the program requests the eleventh page (0260), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the tenth page.
When the program requests the twelfth page (0320), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the eleventh page.
When the program requests the thirteenth page (0370), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the twelfth
Given:
Page size = 100 records
Free main memory frame = 1
Address Sequence:
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370
Calculation:
Initially, the main memory is empty. When the program starts executing, it requests the first page, which causes a page fault because the page is not in the main memory. The operating system then loads the page into the main memory.
When the program requests the second page (0200), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the first page.
Similarly, when the program requests the third page (0430), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the second page.
When the program requests the fourth page (0499), it also causes a page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the third page.
When the program requests the fifth page (0510), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the fourth page.
When the program requests the sixth page (0530), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the fifth page.
When the program requests the seventh page (0560), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the sixth page.
When the program requests the eighth page (0120), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the seventh page.
When the program requests the ninth page (0220), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the eighth page.
When the program requests the tenth page (0240), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the ninth page.
When the program requests the eleventh page (0260), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the tenth page.
When the program requests the twelfth page (0320), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the eleventh page.
When the program requests the thirteenth page (0370), it causes another page fault because the page is not in the main memory. The operating system then loads the page into the main memory, replacing the twelfth
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The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. What is the number of page faults?0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370a)13b)8c)7d)10Correct answer is option 'C'. Can you explain this answer?
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The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. What is the number of page faults?0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370a)13b)8c)7d)10Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. What is the number of page faults?0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370a)13b)8c)7d)10Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. What is the number of page faults?0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370a)13b)8c)7d)10Correct answer is option 'C'. Can you explain this answer?.
The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. What is the number of page faults?0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370a)13b)8c)7d)10Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. What is the number of page faults?0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370a)13b)8c)7d)10Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. What is the number of page faults?0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370a)13b)8c)7d)10Correct answer is option 'C'. Can you explain this answer?.
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